Find the equation of the line which is at a distance of 3 units from the origin such that $\tan\alpha = \frac{5}{12}$, where $\alpha$ is the acute angle which this perpendicular makes with the positive direction of the x-axis.

> [!TLDR] > Since $\alpha$ is acute with $\tan\alpha=5/12$, use the 5-12-13 right triangle to get $\sin\alpha,\cos\alpha$, then substitute into the normal form $x\cos\alpha+y\sin\alpha=p$. > [!FORMULA] > Normal form: $x\cos\alpha+y\sin\alpha=p$. For acute $\alpha$ with $\tan\alpha=\dfrac{\text{opposite}}{\text{adjacent}}$, use the Pythagorean triple to find $\sin\alpha,\cos\alpha$ (both positive in the first quadrant). 1. **Given data.** $$p = 3, \qquad \tan\alpha = \frac{5}{12}, \quad \alpha \text{ acute}$$ 2. **Build the right triangle.** With opposite side $=5$ and adjacent side $=12$, the hypotenuse is $$\sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13$$ 3. **Read off $\sin\alpha$ and $\cos\alpha$** (both positive since $\alpha$ is acute, i.e. in the first quadrant): $$\sin\alpha = \frac{5}{13}, \qquad \cos\alpha = \frac{12}{13}$$ 4. **Substitute into the normal form** $x\cos\alpha+y\sin\alpha=p$: $$x\cdot\frac{12}{13} + y\cdot\frac{5}{13} = 3$$ …

Find the equation of the line which is at a distance of 3 units from the origin such that tanα = (5)/(12), whe

Exercise 12.1 · Q5

\tan\alpha = \frac{5}{12}

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Concept understanding — Straight Line

The Intuition: What Is a Straight Line?

Before any formula, think about what a straight line does. If you walk from your home to school, the shortest path is a straight line. If you stretch a thread tight between two pins, it forms a straight line. The key property is no turning — the direction never changes.

In coordinate geometry, we describe this mathematically. A straight line is the set of all points $(x, y)$ that satisfy a simple rule: the slope (steepness) between any two points on the line is constant.

The Precise Definition

A straight line in the $xy$ -plane is the graph of a linear equation in two variables. The most general form is:

Ax + By + C = 0

where $A$ , $B$ , and $C$ are real numbers, and $A$ and $B$ are not both zero.

Important

Every straight line corresponds to a linear equation, and every linear equation (with $x$ and $y$ ) graphs as a straight line. This is the fundamental link between algebra and geometry.

The Slope: Why It Matters

The slope $m$ measures how steep the line is. It is the ratio of the vertical change to the horizontal change between any two points $(x_1, y_1)$ and $(x_2, y_2)$ on the line:

m = \frac{y_2 - y_1}{x_2 - x_1}

If the line is horizontal, $m = 0$ . If it is vertical, the slope is undefined (division by zero). A positive slope means the line rises as you move right; a negative slope means it falls.

Tip

For a vertical line, the equation is simply $x = \text{constant}$ . For a horizontal line, it is $y = \text{constant}$ .

Common Forms of the Equation

You will encounter these forms frequently. Each is useful in different situations.

1. Slope-Intercept Form

y = mx + c

Here $m$ is the slope and $c$ is the $y$ -intercept — the point where the line crosses the $y$ -axis. This is the most intuitive form: you see the slope and the starting height immediately.

2. Point-Slope Form

y - y_1 = m(x - x_1)

Use this when you know the slope $m$ and one point $(x_1, y_1)$ on the line.

3. Two-Point Form

\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}

Use this when you know two distinct points on the line.

4. Intercept Form

\frac{x}{a} + \frac{y}{b} = 1

Here $a$ is the $x$ -intercept and $b$ is the $y$ -intercept. This form is convenient when the line cuts the axes at known points.

Ax + By + C = 0

Special Cases to Watch For

Line through the origin: $y = mx$ (no constant term).
Horizontal line: $y = k$ (slope $0$ ).
Vertical line: $x = k$ (slope undefined). …

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Since $\alpha$ is acute with $\tan\alpha=5/12$ , use the 5-12-13 right triangle to get $\sin\alpha,\cos\alpha$ , then substitute into the normal form $x\cos\alpha+y\sin\alpha=p$ .

x\cos\alpha+y\sin\alpha=p

Given data.

p = 3, \qquad \tan\alpha = \frac{5}{12}, \quad \alpha \text{ acute}

Build the right triangle. With opposite side $=5$ and adjacent side $=12$ , the hypotenuse is

\sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13

Read off $\sin\alpha$ and $\cos\alpha$ (both positive since $\alpha$ is acute, i.e. in the first quadrant):

\sin\alpha = \frac{5}{13}, \qquad \cos\alpha = \frac{12}{13}

Substitute into the normal form $x\cos\alpha+y\sin\alpha=p$ : $x\cdot\frac{12}{13} + y\cdot\frac{5}{13} = 3$ …

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Q.Find the equation of the line which is at a distance of 3 units from the origin such that tan⁡α=512\tan\alpha = \frac{5}{12}tanα=125​, where α\alphaα is the acute angle which this perpendicular makes with the positive direction of the x-axis.

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Q.Find the equation of the line which is at a distance of 3 units from the origin such that $\tan\alpha = \frac{5}{12}$ , where $\alpha$ is the acute angle which this perpendicular makes with the positive direction of the x-axis.

Unlock everything free for 14 days